#### Question

In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:

(ii) angles APB = 90°

#### Solution

ii) Now in Δ ATP ,

∴ `∠`TAP = `∠`TPA

Similarly in Δ BTP,`∠`TBP = `∠`TPB

Adding,

`∠`TAP +`∠`TBP =`∠`APB

But

∴ TAP + `∠`TBP + `∠`APB =180°

⇒ `∠`APB = `∠`TAP + `∠`TBP =90°

Is there an error in this question or solution?

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In the Given Figure, Two Circles Touch Each Other Externally at Point P. Ab is the Direct Common Tangent of These Circles. Prove That: (Ii) Angles Apb = 90° Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.

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